Converging Lens Homework: Finding Image Distance with Thin Lens Equation"

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The discussion revolves around calculating the image distance for a new converging lens with a focal length of 32.0 cm, given an initial lens with a focal length of 11.0 cm that produced a real image at 25.0 cm. Using the thin lens equation, the object distance was determined to be 0.196 m for the first lens. For the second lens, the calculations yielded an image distance of -0.506 m, indicating a virtual image. Participants noted the importance of maintaining significant figures in calculations to improve accuracy. The final consensus is that the image produced by the second lens is virtual.
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Homework Statement



An object is in front of a converging lens with a focal length of 11.0 cm. The lens
produces a real image located 25.0 cm from the lens. If the lens is replaced by a new lens
(in the same position) that has a focal length of 32.0 cm, what is the image distance of the
new lens? Is the image real or virtual?

Homework Equations



Thin lens equation
1/f=1/d0+1/di

The Attempt at a Solution



lens 1:

1/+11*10-2=1/dO+1/+25.0*10-2

dO=0.196m

lens2:

1/+32.0*10-2=1/0.196m+1/di

di=-0.506m

Virtual image
 
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aChordate said:


lens 1:

1/+11*10-2=1/dO+1/+25.0*10-2

dO=0.196m

lens2:

1/+32.0*10-2=1/0.196m+1/di

di=-0.506m

Virtual image


Your approach is correct.

Your answer is pretty close, but there are some rounding errors. You might want to try again, but keep more significant figures in the intermediate steps.
 
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I got -.509 on the second try. Hopefully that is correct.
 
aChordate said:
I got -.509 on the second try. Hopefully that is correct.
Yup. :approve: (Don't forget your units though.)
 
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